Minimum Difference Representations of Graphs - Semantic Scholar

Minimum Difference Representations of Graphs J´ozsef Balogh∗

Noah Prince†

February 8, 2007

Abstract Define a k-minimum-difference-representation (k-MDR) of a graph G to be a family of sets {S(v) : v ∈ V (G)} such that u and v are adjacent in G if and only if min{|S(u) − S(v)|, |S(v) − S(u)|} ≥ k. Define ρmin (G) to be the smallest k for which G has a k-MDR. In this note, we show that {ρmin (G)} is unbounded. In particular, we prove that for every k there is an n0 such that for n > n0 ‘almost all’ graphs of order n satisfy ρmin (G) > k. As our main tool, we prove a Ramsey-type result on traces of hypergraphs.

1

Introduction

A representation of a graph is an assignment of sets of a given type (intervals, convex sets, subsets of N, etc.) to the vertices of G such that u, v ∈ V (G) are adjacent if and only if S(u) and S(v) satisfy a given condition (are disjoint, intersect, etc.). Many different kinds of graph representations have been studied extensively (see, for example, [5, 8, 10]). In [7], three new definitions of graph representations were introduced. With each vertex v of G, a finite set S(v) is associated. In the first model, u and v are adjacent if and only if (|S(u) − S(v)| + |S(v) − S(u)|) /2 ≥ k, in the second, iff max{|S(u) − S(v)|, |S(v) − S(u)|} ≥ k, and in the third, iff min{|S(u) − S(v)|, |S(v) − S(u)|} ≥ k. ∗ University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA; email: [email protected], research supported in part by NSF grants DMS-0302804, DMS-0603769 and DMS-0600303, UIUC Campus Research Board 06139 and 07048, and OTKA 049398. † University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA; email: [email protected], research supported in part by UIUC Campus Research Board 07048.

1

Let ρavg (G), ρmax (G), and ρmin (G) be the smallest k for which G has such a representation in the first, second, and third model, respectively. In [7] it was shown that each of these parameters exists and is at most n − 1, where n is the order of the graph G. It was also shown that the first two parameters could be large, namely, at least O (log n). F¨ uredi [9] showed that, for almost all n-vertex graphs, all three parameters are at most O(n/ log n) and ρmax (G) = Ω(n/ log n). However, no non-trivial lower bounds on ρmin (G) have been proven, and in [7] it was noted that no graphs were known for which ρmin (G) > 2. In this short paper, we will show that {ρmin (G)} is unbounded by demonstrating that, for any k ∈ N, almost every ‘large’ graph G satisfies ρmin (G) > k. Theorem 1 Let k be a positive integer. Then G ∈ Gn,1/2 satisfies ρmin (G) > k with high probability as n → ∞. Our method is as follows. First, we establish a Ramsey-type result on traces of hypergraphs (Theorem 3) saying that a large enough family of sets must contain one of four configurations. When we apply this to the sets of an optimal k-minimum-difference-representation (k-MDR) of G, that is, {S(v) : v ∈ V (G)}, we can immediately rule out three of these (see Lemma 6). The remainder of the proof is dedicated to ruling out the fourth configuration. Before we begin, let us recall some definitions and notation. A set system is a family of finite sets. For two sets A and B, A ⊆ B means A is a subset of B, and A ⊂ B means that A is a proper subset of B. A set system H is an antichain if no A, B ∈ H exist with A ⊂ B. If G is a graph and X ⊆ V (G), then NG (X) is the set of vertices in V − X which have a neighbor in X. We use n Gn,1/2 to denote the probability space of all 2( 2 ) graphs on n labelled vertices chosen with uniform distribution. Throughout the paper, all logarithms will have base 2.

2

Traces

In this section we prove a Ramsey-type theorem on set systems. Define a trace of a set system H on X to be the system HX = {E ∩ X : E ∈ H}. We say that HX is induced by X, and the trace sets are the sets E ∩ X. We say E is an ancestor in H of E ∩ X. We also consider any subsystem of HX to be a trace of H induced by X. Define an (`, k)-star to be a set system {Ai : 1 ≤ i ≤ `} whose elements are pairwise disjoint and all have S cardinality k. An (`, k)-costar is a set system {Ai : 1 ≤ i ≤ `} where, if A = i Ai , then {A − Ai : 1 ≤ i ≤ `} is an (`, k)-star. An `-star or `-costar is simply a (`, 1)-star or (`, 1)-costar, respectively. An m-double-chain is a system {{vi , . . . , vi+m−1 } : 1 ≤ i ≤ m}. Balogh and Bollob´as [2] proved the following result on traces of set systems. `

Theorem 2 Let ` and m be positive integers and let g(`, m) = (2m)` 29 . Then any antichain H with |H| ≥ g(`, m) contains either an `-star, an `-costar, or an m-double-chain as a trace. 2

A short proof of Theorem 2, along with other similar results, appeared in [6]. Recently, this result has seen many applications, for example in [1], [3] and [4]. Our new theorem involves a fourth kind of set system. Define an (r, p, q)system to be a set system [ [ Lj : 1 ≤ i ≤ r} Gj ∪ {K ∪ Fi ∪ j≤i

j6=i

such that (1) K is the kernel of the system, (2) K and the 3r sets Fi , Gi , and Li for 1 ≤ i ≤ r are pairwise disjoint, (3) |Fi | = p and |Gi | = q for all 1 ≤ i ≤ r. .................................................................................................................................... ......................................... .......... ...... .......... ....... ...... ..... ....... ...... .... .... ..... ..... .... .... .... .... ... ... . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... ....... . ............................ ............................. ... . ... .. .... .. ...... ...... ...... .... .... .... ..... . . . . ... . . . . . . . . . .... . . ... ... ... . . . ... ... ... .. ... ... ... ... ... ... ... ... ... . . . .... .... . ... ... .. ... .. .. .. ... ... ... ... . . . . ... .... . .... . . . ... ... ... ... ... .. .. .. .. ... . . . . . . . . . ... . . . . .... .... ... ... ... ... ... .. ... ... ... 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... . ... .. ... .... .... ... . . . .... . . . .... . ... ... ... ... .. ... .. ... ... ... ... .... ... ... ... ... . . . . ..... .... .... .... .... ... .. ... .. .. .. . . . ... . . . . . . . . . . . . . . . . .... ... . . . . . ......... ........ ......... ........ ......... ........ .... ... ............ ............ ............ .. ... ..... . . . . .... ..... . ......... .. . ....... ...................... ... .. .. ... ............................. .. ...................................................... .. ... ... ............ ....... .. ... .. ..... ....... ... ... .. ... ... ..... ............................. ............................. ............................. ... ... .. ... .. ... ...... ...... ...... .... .... .... . . . . . . . . ..... . . ..... .. . . . ... ... ... ... . .. .. .. .. .. . ... ... ... ... .. ... ... ... ... ... .. .. .. .. ... .. .... .... ..... ..... .... ..... ..... . ... ... .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... . . ... . .... . .. . . . . . . . . . . . . . . . . . . . . ... ...... ....... ... . ........ ... .. ............................. .... .. ............................ ..... ..... ......................... .. .. ...... ..... ....... .. ... ... .......... ....... ........... .. .................................... ................................................. ... ... .................................. ............. ... .. ... .................. ...... ....... ... ... ......... . . . . . . . . . . . . . . . ..... ... ... . .. ............................ ..... ............................ ............................ ... ... ... ....... ....... ....... ... ... .... .... .... . ... .. .... .... .... ... ... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... .. .. .. ... ... .. ... .. ... ... ... ... ..... ..... ..... ... ... .. . ... .. ... . . . . . . . . . . . . ... . ... ... ... ... ... ... ... .... ... ... .... .... . ... ... .... . ... . . . . . . . . . ... ..... .. ...... ...... ... ... . ... ... ... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................... ................... ................... ... .. ... . . .. ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... .. .. .. ... ... ... ... ... ... ... ... ... .. .. .. . . . . . . . ... ... ... . . ... ... ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... .. ... ... ... ... ... ... ... ... ... ... .. .. . . . . . . . ... ... ... ... ... ... ... ... ... .. .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. .. ... ... . . . . . . . . . . ... ... ... . . ... ... ... ... ... ... ........................... ........................... ............................ ... .. ... ... ....... ....... .... ....... .... .... ... ... ... ... ... ... ... ... .... .... ... .... .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. .. ... ... .. ... ... .. .. . . . . . . . . . . ... . . . . . ... ... ... ... ... .. ... .. .. .... .. .. . . . . . . . . . . . . . . . . ... . . ... ... ... ... ... ... ... ... ... .... ... .. .... . . . . . . . . ... . . . . . . . . . . . . . . . . . ...... ...... ....... ... ...... ...... ... .. .. ............................. ............................. ....... . . . . . . . .............. . . . . . . . . . ... . . . . . . . . . . . . . .... . .. .... .... ... .... ..... ..... ... ..... ...... .... ...... ...... ........ ..... ........... ................................................. .............................

K

F1

G1

L1

. .. .

. .. .

. .. .

Fi−1

Gi−1

Li−1

Fi

Gi

Li

Fi+1

Gi+1

Li+1

.. ..

.. ..

.. ..

Fr

Gr

Lr

Figure 1. A general element of an (r, p, q)-system

Define h1 (`, m, r) = g(`, m) and hk (`, m, r) = g(r 2 · hk−1 (`, m, r), m) for k > 1. Theorem 3 Let `, k, m, and r be positive integers. Then every antichain H with |H| ≥ h2k (`, m, r) contains an (`, k)-star, an (`, k)-costar, or an m-double3

chain as a trace or contains an (r, p, q)-system with p, q ≤ k as a subsystem. Furthermore, we may insist that the (r, p, q)-system have Li = ∅ for all i or L1 = ∅ and Li 6= ∅ for all i ≥ 2. Throughout the proof, we use the term collection to mean a set whose elements may occur with multiplicity greater than 1. Proof. If H has an m-double-chain as a trace, we are done. Otherwise, implement the following algorithm: (Step 1) Initialize p = q = 0, i = 1 and H1 = H. (Step 2) Let di = r2 · h2k−i (`, m, r). If Hi contains no di -star trace, move to Step 3. Otherwise, let T1i , . . . , Tdii be the trace sets of a di -star trace, increment p by 1, and call the set inducing the star Bp . Let Aij be an ancestor in Hi of Tji , and define Hi+1 to be the set system {Aij − Bp : 1 ≤ j ≤ di }. Increment i by 1. If p = k, stop and output the ancestors in H of the Tji ’s; these sets contain an (`, k)-star trace. Otherwise repeat Step 2. (Step 3) If Hi contains no di -costar trace, move to Step 4. Otherwise, let T1i , . . . , Tdii be the trace sets of a di -costar trace, increment q by 1, and call the set inducing the costar Cq . Let Aij be an ancestor in Hi of Tji , and define Hi+1 = {Aij − Cq : 1 ≤ j ≤ di }. Increment i by 1. If q = k, stop and output the ancestors in H of the Tji ’s; these sets contain an (`, k)-costar trace. Otherwise return to Step 2. (Step 4) Find D1 , D2 , . . . , Dr among the sets T1i , . . . , Tdii such that D1 = D2 = . . . = Dr or D1 ⊂ D2 ⊂ . . . ⊂ Dr , and let D ⊂ H be the collection of ancestors in H of Di ’s. Stop and output D; these sets contain an (r, p, q)-system with p, q ≤ k. We claim this algorithm outputs an (`, k)-star, an (`, k)-costar, or an (r, p, q)system with p, q ≤ k. First, we show the algorithm is well-defined through 2k − 1 steps. Since i ≤ 2k − 1, Steps 2 and 3 are well-defined. Therefore, the only thing that must be checked is that when we enter Step 4, we can always find r sets all equal or ordered linearly by proper inclusion. Suppose that we enter Step 4 on the ith turn. If any set appears with multiplicity r among the Tji ’s, then we can take D to be r distinct ancestors in H of this set. Otherwise, there are at least di /r = r · h2k−i (`, m, r) distinct sets among the Tji ’s, so by Dilworth’s Theorem there is a chain of length r or an antichain of size h2k−i (`, m, r) among the Tji ’s. In the first case we are clearly done. In the second, this antichain contains a di -star, a di -costar, or an m-double-chain as a trace by Theorem 2. The first two cases are impossible since we did not complete Steps 2 and 3, and the third is impossible by assumption. Hence such D1 , D2 , . . . , Dr exist. 4

Second, the algorithm always terminates with i ≤ 2k − 1 since p + q increases by 1 after every complete implementation of Step 2 or 3 and we either stop when max(p, q) = k, or we stop after the first implementation of Step 4. Finally, we show the algorithm produces what we want. If the algorithm terminates during the ith turn because p = k, then we found a star trace T1i , . . . , Tdii where di ≥ `. Note that the Tji ∩ Bn ’s are disjoint and have cardinality 1 for all 1 ≤ j ≤ di and 1 ≤ n ≤ p = k. Letting Aij be an ancestor in H of Tji , we see Ai1 , . . . , Ai` form an (`, k)-star induced by p [

n=1

Bn

!

∩

` [

i=1

Aij

!

.

The case when the algorithm terminates because q = k is similar. Now, if the algorithm terminated after an implementation of Step 4, let A1 , . . . , Ar be distinct such that Ai is an ancestor in H of Di for 1 ≤ i ≤ r. Then the sets Ai ∩Bn and Cs −Ai are pairwise disjoint Sp and have cardinality Sq 1 for all 1 ≤ i ≤ r, 1 ≤ n ≤ p, and 1 ≤ s ≤ q. Let Fi = n=1 (Ai ∩Bn ), Gi = S ), L1 = ∅ s=1 (Cs −AiS r q and Li = Di − Di−1 for 2 ≤ i ≤ r, and let K = D1 ∪ ( n=1 Cn − i=1 Ai ). Then we have, for all i, that [ [ Ai = K ∪ F i ∪ Gj ∪ Lj , j6=i

j≤i

and 0 ≤ |Fi | = p ≤ k and 0 ≤ |Gi | = q ≤ k, so the Ai ’s form an (r, p, q)-system with kernel K. Note that if the Di ’s were all the same, then Li = ∅ for all i, whereas if they were all different, L1 = ∅ and Li 6= ∅ for all i ≥ 2. 

3

Proof of the main result

In this section, we prove our main result, namely, Theorem 1. Before we begin, we must recall two basic facts about random graphs. Fact 4 Almost every G ∈ Gn,1/2 has an independent set on log n vertices. Fact 5 Let G ∈ Gn,1/2 and let S ⊆ V (G) with |S| = o(log n). If S is the disjoint union of A and B, then with high probability there is a vertex v such that v is adjacent to every vertex in A and to no vertex in B. We begin with an easy observation. Lemma 6 Let H be a set system satisfying min{|Si − Sj |, |Sj − Si |} < k for all Si , Sj ∈ H. Then H has no (2, k)-star, (2, k)-costar, or (k + 1)-double-chain as a trace.

5

Proof. Suppose X induces a (2, k)-star trace with trace sets T1 and T2 . Let Si be an ancestor in H of Ti . Then |Si − S3−i | ≥ |(Si − S3−i ) ∩ X| = |Ti − T3−i | = k for i = 1, 2, so min{|S1 − S2 |, |S2 − S1 |} ≥ k, a contradiction. The statement for (2, k)-costars follows since a (2, k)-costar is a (2, k)-star. Now suppose H contains a (k + 1)-double-chain trace induced by X with trace set T1 , . . . , Tk+1 . By the definition of the double-chain, |Tk+1 − T1 | = |T1 − Tk+1 | = k, so if Si is an ancestor in H of Ti , then as above min{|S1 − Sk+1 |, |Sk+1 − S1 |} ≥ k, a contradiction. Hence H has no (k + 1)-double-chain as a trace.  Proof of Theorem 1. Suppose, toward a contradiction, that G has a k-MDR; call it S = {S(v) : v ∈ V (G)}. Note that S(u) = S(v) iff u = v, since with high probability no two distinct vertices of G have the same neighborhood (by Fact 5). By Fact 4, with p high probability, G contains an independent set I = {v1 , . . . , v|I| } with |I| = log(n). Let SI = {S(vi ) : vi ∈ I}. Suppose SI contains a chain of length 3. Without loss of generality S(v1 ) ⊂ S(v2 ) ⊂ S(v3 ). By Fact 5 there is a vertex x such that xv1 , xv3 ∈ E(G) and xv2 6∈ E(G). The fact that xv1 ∈ E(G) tells us |S(v1 ) − S(x)| ≥ k, so |S(v2 ) − S(x)| ≥ k as well. Similarly, |S(x) − S(v2 )| ≥ |S(x) − S(v3 )| ≥ k. But then xv2 ∈ E(G), a contradiction. √ Therefore by Dilworth’s Theorem, there is an antichain in SI of size log n/2. Without loss of generality, call this antichain p S ∗ = {S(vi ) : 1 ≤ i ≤ log n/2}. Note that, for all S1 , S2 ∈ S ∗ ,√min{|S1 − S2 |, |S2 − S1 |} < k. Take n large enough that log n/2 ≥ h2k (2, k + 1, r) with r = 20k 4 (2k + 1). By Theorem 3, S ∗ contains a (2, k)-star, a (2, k)-costar, a (k + 1)-double-chain, or an (r, p, q)-system with p, q ≤ k, L1 empty, and the Li ’s with i ≥ 2 all empty or all nonempty. By Lemma 6, only the fourth case is possible. Call the sets of this (r, p, q)-system S(v1 ), . . . , S(vr ). Let Vt = {v20k4 t + 1, . . . , v20k4 (t+1) } for 0 ≤ t ≤ 2k. Let K, Fi , Gi , and Li be as in the definition of (r, p, q)-system, so that, for 1 ≤ i ≤ r, [ [ Lj . Gj ∪ S(vi ) = K ∪ Fi ∪ j6=i

j≤i

With high probability, we can find a set of vertices X = {x1 , . . . , x4k2 } (applying Fact 5 for each xi ) such that xi is adjacent to vj iff, for some 0 ≤ t ≤ 2k, 5k 2 (i − 1) + 1 + 20k 4 t ≤ j ≤ 5k 2 i + 20k 4 t 6

(from here on we only use the vj ’s with j ≤ r) and X is an independent set in G. Observe the following two properties: (1) for each i and t, |NG (xi ) ∩ Vt | = 5k 2 , (2) for each i, j, and t with i 6= j, |NG (xi ) ∩ NG (xj ) ∩ Vt | = 0. We consider the two cases depending on whether all the Lj ’s with j ≥ 2 are empty or nonempty. CASE 1: Lj = ∅ for all 1 ≤ j ≤ r. For the proof of this case, we work only with vj ’s in V0 . Fix fi,j = |Fj ∩ S(xi )| and gi,j = |Gj − S(xi S )|, and let P P some xi ∈ X. Define fi = j fi,j and gi = j gi,j . Let ei = |K − S(xi )| and ci = |S(xi ) − j S(vj )|. Then, for any vj ∈ V0 , |S(vj ) − S(xi )| = p − fi,j + ei + gi − gi,j

(1)

|S(xi ) − S(vj )| = fi − fi,j + ci + q − gi,j .

(2)

and

Consider a vj not adjacent to xi . By the definition of k-MDR, either (i) |S(vj ) − S(xi )| < k or (ii) |S(xi ) − S(vj )| < k. We claim there are not both vj1 which satisfies (i) and vj2 which satisfies (ii). Indeed, since (i) holds for vj1 , gi ≤ |S(vj1 ) − S(xi )| + gi,j < 2k, and, similarly, since (ii) holds for vj2 , we have fi ≤ |S(xi ) − S(vj2 )| + fi,j < 2k. Hence, for a fixed i, fewer than 2k fi,j ’s and fewer than 2k gi,j ’s are nonzero (this includes vj ’s adjacent to xi ). Then we can find vertices vj3 , vj4 with fi,j3 = gi,j3 = fi,j4 = gi,j4 = 0 such that xi is adjacent to vj3 but not to vj4 (here we use Property (1) from above). But then |S(vj4 ) − S(xi )| = |S(vj3 ) − S(xi )| ≥ k |S(xi ) − S(vj4 )| = |S(xi ) − S(vj3 )| ≥ k,

contradicting that xvj4 6∈ E(G). Therefore, for each xi , either

Alternative 1: each vj not adjacent to xi satisfies (i) and not (ii), or Alternative 2: each vj not adjacent to xi satisfies (ii) and not (i).

Now, by the pigeonhole principle, there are either at least 2k xi ’s such that Alternative 1 holds or at least 2k xi ’s such that Alternative 2 holds (we could in fact have 2k 2 , but 2k suffices for this case). Assume the former, and without loss of generality let these vertices be X0 = {x1 , . . . , x2k }. Delete from V0 all vj ’s for which there is an xi ∈ X0 such that Gj 6⊆ S(xi ), and let V 0 be the subset of V0 that remains. Since (i) holds for all xi ∈ X0 , at most 2k vj ’s satisfy Gj 6⊆ S(xi ) for each xi , so |V0 − V 0 | ≤ 2k|X0 | = 4k 2 . 7

Therefore |NG (xi ) ∩ V 0 | ≥ k 2 for all xi ∈ X0 . Fix some xi ∈ X0 . Since Alternative 1 holds for xi , we see xi vj 6∈ E(G) if and only if |S(vj )−S(xi )| < k. But, for all vj ∈ V 0 , |S(vj )−S(xi )| = p−fi,j +ei +gi (since gi,j = 0, by definition of V 0 ), and among the terms on the right, only fi,j depends on j. Therefore there is a threshold t(xi ) such that if vj ∈ V 0 , then xi is adjacent to vj iff fi,j ≤ t(xi ). Since each xi ∈ X0 is nonadjacent to some vj ∈ V 0 , we deduce that t(xi ) < p ≤ k. Since |X0 | > k, there are two elements of X0 , without loss of generality x1 and x2 , with t(x1 ) = t(x2 ). Now, for all j such that vj ∈ V 0 ∩ NG (x1 ) (so vj 6∈ NG (x2 )), |Fj ∩ S(x1 )| ≤ t(x1 ) = t(x2 ) < |Fj ∩ S(x2 )|, so there is an element of Fj in S(x2 ) and not in S(x1 ). Therefore, |S(x2 ) − S(x1 )| ≥ |V 0 ∩ NG (x1 )| ≥ k 2 .

Similarly, |S(x1 ) − S(x2 )| ≥ |V 0 ∩ NG (x2 )| ≥ k 2 . Hence x1 x2 ∈ E(G), a contradiction. The case when at least 2k xi ’s satisfy Alternative 2 is similar. Call the set of these vertices X0 . We find a V 0 ⊆ V0 such that for all xi ∈ X0 and all vj ∈ V 0 , S(xi ) ∩ Fj = ∅ (that is, fi,j = 0) and |V 0 ∩ NG (xi )| ≥ k 2 . Then for each xi , there is a threshold t(xi ) such that xi is adjacent to vj ∈ V 0 iff gi,j ≤ t(xi ). We then find two vertices x1 and x2 with the same threshold, and necessarily min{|S(x1 ) − S(x2 )|, |S(x2 ) − S(x1 )|} ≥ k 2 , so x1 and x2 are adjacent in G, a contradiction. This completes Case 1.

CASE 2: L1 = ∅ and Lj 6= ∅ for all j ≥ 2. Define [ Mt = Lj vj ∈Vt

for 0 ≤ t ≤ 2k. Now, if S(xi ) contains an element of Mt for each t ≥ k + 1 and Mt −S(xi ) is nonempty for each t ≤ k−1, then |S(xi )−S(vj )| and |S(vj )−S(xi )| are at least k for all vj ∈ Vk , contradicting that xi has a nonneighbor in Vk . Therefore, for each xi , one of these cannot happen. Suppose first that for at least 2k 2 xi ’s, it is not the case that S(xi ) contains an element of Mt for each t ≥ k + 1, so there is a ti ≥ k + 1 such that S(xi ) contains no element of Mti . Then there is a set X0 of 2k xSi ’s each with the same ti , say ti = m for these xi ’s. Then, if we define bi = | tj

S where si = | t>m Mt ∩ S(xi )|. Fix xi ∈ X0 and let vj ∈ V ∗ be a nonneighbor of xi , so necessarily |S(xi ) − S(vj )| < k. Since gi,j ≤ q, it follows that, fi ≤ |S(xi ) − S(vj )| + fi,j ≤ 2k − 1. Therefore, for each i, at most 2k − 1 fi,j ’s are nonzero. Delete from V ∗ each vj for which there is an xi ∈ X0 such that fi,j 6= 0, and call the remaining vj ’s V 0 . Then |V ∗ − V 0 | ≤ (2k − 1)|X0 | ≤ 4k 2 − 2k, so each xi ∈ X0 has at least 5k 2 − |Vm − V ∗ | − (4k 2 − 2k) ≥ k 2 neighbors in V 0 . Since fi,j = 0 for all xi ∈ X0 and vj ∈ V 0 , we see, for a fixed i, |S(xi ) − S(vj )| depends only on gi,j . Thus there is a threshold t(xi ) ≤ k such that vj is adjacent to xi iff gi,j ≤ t(xi ). Since |X0 | = 2k, there are two elements of X0 , say x1 and x2 , with the same threshold. As in the proof of Case 1, it can be proved that x1 and x2 are adjacent, a contradiction. Finally, we are left with the case that there are at least 2k 2 xi ’s such that Mt − S(xi ) is empty for some t ≤ k − 1. This case is similar to the previous case, except that the roles of |S(xi ) − S(vj )| and |S(vj ) − S(xi )| are reversed; we skip the details.  Acknowledgement. We are grateful for Zolt´an F¨ uredi for drawing attention to this problem.

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